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Geometry Question Quant 2nd Ed #102 pg. 116

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by OneTwoThreeFour » Fri Apr 08, 2011 9:21 pm
Please tell me why my calculations are wrong. Thanks!

In the coordinate plane, line k passes through the origin and has slope 2. If points (3,y) and (x,4) are on line k, then x+y=

(A) 3.5
(B) 7
(C) 8
(D) 10
(E) 14

Answer: C

My calculations:

Since slope is 2 and k passes through the origin, the y-intercept is zero. Thus y=2x. Since we know that points (3,y) and (x,4) are on line k, then that means,

(4-y)/(x-3) =2.
10=2x+y
Given that y=2x, I set the two equations, 10=2x+y and y=2x, equal to each other and got x=2.5 and y=5. Clearly this isn't the right answer. Why are my calculations wrong?

Thanks for your help!
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by vineeshp » Fri Apr 08, 2011 10:40 pm
(4-y)/(x-3) =2.
10=2x+y
Given that y=2x, I set the two equations, 10=2x+y and y=2x, equal to each other and got x=2.5 and y=5. Clearly this isn't the right answer. Why are my calculations wrong?
You have calculated the slope and equated it to the line. Not correct.

There is no way 10=2x+y represents the same line. In fact you will never have two equations for one line.

The equation of the line is
y - y1 = m ( x-x1)
and m = y2 - y1 / x2 - x1


The simpler way to solve this is ... Once you arrive at y = 2x
(3,y)
y = 2x. x is given as 3 so y = 6.
In the other equation
y= 2x = 4. x=2.

x +y = 6 + 2 = 8.
Vineesh,
Just telling you what I know and think. I am not the expert. :)
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by GMATGuruNY » Sat Apr 09, 2011 1:43 am
OneTwoThreeFour wrote:Please tell me why my calculations are wrong. Thanks!

In the coordinate plane, line k passes through the origin and has slope 2. If points (3,y) and (x,4) are on line k, then x+y=

(A) 3.5
(B) 7
(C) 8
(D) 10
(E) 14

Answer: C
No need to set up complicated equations or to determine the equation of the line. Just use the slope formula.

Slope = (y₂ - y�)/(x₂ - x�).
Given slope = 2.

(0,0) and (3,y) are on the line.
(y-0)/(3-0) = 2.
y/3 = 2.
y=6.

(0,0) and (x,4) are on the line.
(4-0)/(x-0) = 2.
4/x = 2.
x=2.

x+y = 2+6 = 8.

The correct answer is C.
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by OneTwoThreeFour » Sat Apr 09, 2011 7:56 am
Thanks everyone!
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